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Calculus I                                                          Quiz #4                            February 24, 2004

Name____________________                   R.  Hammack                              Score ______
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(1)
(a)  cos((5π)/4) =-1/2^(1/2)

(b)  tan((5π)/4) =sin((5π)/4)/cos ((5π)/4) = -1/2^(1/2)/-1/2^(1/2) = 1

(c) sec((5π)/4) =1/cos ((5π)/4) = 1/-1/2^(1/2) = -2^(1/2)



(2)  Find the domain of the function f(x) = x^(1/2)/(1 + 2cos(x))

First, x can't be negative, for that would involve the square root of a negative number.
Second, x can't be such that 1+2cos(x) = 0, for that would result in division by zero.

The equation 1+2cos(x) = 0 reduces to cos(x) = -1/2,
so we are looking out for values of x that make cos(x) = -1/2.  
Looking at the unit circle, you can see that x = (2π)/3and x = (4π)/3make cos(x) = -1/2.
[Graphics:HTMLFiles/quiz4_10.gif]
Of course, you could add any multiple of 2π to these values, and still cos(x) = -1/2,
making the denominator in f(x) zero.

Conclusion:  The domain is {x | x≥0,    x ≠ (2π)/3 + 2n π,   x ≠ (4π)/3 + 2n π , where n is an integer }



(3)  In this problem, θ is a number between 0 and π/2 for which csc(θ) = 2^(1/2).

This information says 1/sin(θ) = csc(θ) = 2^(1/2),
so it follows that  sin(θ) = 1/2^(1/2).  From this it follows that θ = π/4, so cos(θ) = 1/2^(1/2).
Then:
(a)  sin(θ) = 1/2^(1/2)
(b) tan(θ) =sin(θ)/cos(θ) = 1/2^(1/2)/1/2^(1/2) = 1